XTR106

Hi Collin,

I just resumed this project and your design file is very useful. I have calculated these values, could you please confirm if they are correct (see datasheet of photodiode I am using)

R1 = 5.36 MOhm     C1 = 2.7pF     Cin = 5009pF     fGBW = 20.413 MHz     R2 = 13.7 kOhm     R3 = 280 Ohm     fp = 580.028 Hz     fFP = 663.145 kHz

I have a few questions:

1) On page 2 of the file,what the goals are based and how did you calculate them e.g. (Gain: 53.6kV/A, -3dB Bandwidth : >1MHz, Phase Margin: >45MHz etc)

2) When I simulate in TINA using my calculated values, I get a good DC transfer function like yours but AC function is not giving me -3dB bandwidht of >1MHz (see attachment). Is it because of too high fGBW that I have and therefore I need to use a different amplifier with higher bandwidth? Could you please help me as I am stuck on this and not moving forward. Thank you

G1963_Datasheet.pdfG1963_Datasheet.pdf

Regards,

Saif

Hi Saif,

Most of the design goals are simply goals we would like to achieve for the design. The theory section of the document describes how to achieve the goals or adjust the design to meet your specific design requirements.

The gain is based on the minimum and maximum output swing of the amplifier and the maximum input current used in the design. The gain is calculated using equation 3 in the design file.

The -3dB bandwidth is determined by the selection of R1 and C1. R1 and C1 form a pole in the frequency response, this pole can be calculated using equation 4 in the design file.

The phase margin is a merit of how stable a circuit is. For a stable circuit it is recommended to have greater than 45 degrees of phase margin. For more information on stability, please see our TI Precision Labs on Op Amp Stability.

The reason the design is not getting >1MHz of bandwidth is because R1=5.36M ohm and C1=2.7pF. This creates a pole in the frequency response at approximately 11kHz (see equation 4 in the design file). To get more bandwidth you will need to decrease R1 which in turn will decrease the gain of the circuit. This is described in section 2.2 and 2.3 of the design file.

-Tim Claycomb

Hi Tim,

If I decrease my R1 value, then it will change my gain as you said and secondly, I calculated all values according to the design reference provided by Collin above so now I am totally confused what to do.
R1 = Vout(max) - Vout(min)/Iin(max) so R1 = 4.9V-0.1V/0.9uA gives you 5.36Mohms so I don't see how can I decrease the R1 value?

Regards,
Saif

Hi Saif,

What will be the frequency of your input signal?

To get the bandwidth up to 1MHz with R1=5.36M ohms the feedback capacitor (C1) will need to decrease to 29.7fF. However, the smallest value capacitor I could find to purchase is 0.1pF. A 0.1pF feedback capacitor would give a cut off frequency of 296kHz and will likely not stabilize the amplifier.

One alternative option would be to select a different photo-diode that can output a larger current. This would allow the value of R1 to decrease. For example the photo-diode used in the design file is SFH213 made by Osram. This photo-diode can output 90uA.

-Tim Claycomb

Hi Tim,

Thanks for replying. For now, I will have to stick with G1963 from Hamamatsu. See the picture below for your information. This is the first time I am designing such a sensor so I am a beginner. I want to start from beginning, if you think I need to use a different amplifier please feel free to tell me as i am only using OP320 because it was used in the reference design file.

So, what would be my goals first of all and how to determine it. I could say I need 1GHz but how to justify it like you said you need > 1MHz for -3dB Bandwidth?

Regards,

Saif Malik

Hi Tim,

Wouldn't it be nice to show how to use a hamamatsu photodiode with TI amplifier and how to design the sensor using datasheet information. My ultimate goal will be to have the sensor as 4-20mA signal where 04mA = 0W/m2 20mA=2000W/m2.

Regards,
Saif Malik

Hi Saif,

The bandwidth of the circuit will depend on the frequency of your input signal. The input signal in your case will be a light source for the photo-diode to detect. For example, in the design file a test to verify performance of the design was to use a 500kHz square wave light source generated from another PCB (TIPD108) as the input. That was the input frequency for that test.

So the bandwidth required for your design is dependent on the rate at which you expect the light source to change. If you expect it to change once every millisecond then you only require 1kHz of bandwidth. If you expect it to change once every 100 microseconds then you only require 10kHz of bandwidth.

-Tim Claycomb